Answer: 60
Step-by-step explanation:
As JH=22 and JK=82, this means HK=60.
The work done (in foot-pounds) in stretching the spring from its natural length to 0.7 feet beyond its natural length is 1.23 foot-pound
<h3>Data obtained from the question</h3>
From the question given above, the following data were obtained:
- Force (F) = 3 pounds
- Extension (e) = 0.6 feet
- Work done (Wd) =?
<h3>How to determine the spring constant</h3>
- Force (F) = 3 pounds
- Extension (e) = 0.6 feet
- Spring constant (K) =?
F = Ke
Divide both sides by e
K = F/ e
K = 3 / 0.6
K = 5 pound/foot
Thus, the spring constant of the spring is 5 pound/foot
<h3>How to determine the work done</h3>
- Spring constant (K) = 5 pound/foot
- Extention (e) = 0.7 feet
- Work done (Wd) =?
Wd = ½Ke²
Wd = ½ × 5 × 0.7²
Wd = 2.5 × 0.49
Wd = 1.23 foot-pound
Therefore, the work done in stretching the spring 0.7 feet is 1.23 foot-pound
Learn more about spring constant:
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It is called a bonding kilyodhuifguegdhegufguygfgerugued
Answer:
Day 2
Step-by-step explanation:
Every 20 days, one jar is full.
To find when it was 1/8 full, we multiply the 20 days by 1/8.
20 x 1/8 = 2
Day 2
Answer:
Step-by-step explanation:
5x + 9y = -11
3x + 9y = -3
5x + 9y = -11
-3x - 9y = 3
2x = -8
x = -4
-12 + 9y = -3
9y = 9
y = 1
(-4, 1)
Answer is C
